FODE_1-10

FODE_1-10 - Tutorial FODE: Ordinary First-Order...

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6. Solutions to Exercises Exercise 1. The antiderivative of 2 x is x 2 : Thus, the general solution is y ( x ) = x 2 + C: The condition tells us 0 =1 2 + C = ) C = ¡ 1 ; or y ( x ) = x 2 ¡ 1 : Exercise 2. Writing ln A insteadof C as the constant of integration, we can write Eqn. 4 as ln j y ln A = ln μ j y j A = x 2 2 = ) y = A exp μ x 2 2 : Exercise 3. y (0)= 2 = ) 2 = A ¢ 1 = ) A = 2 : Thus, y ( x )= 2exp μ x 2 2 : Exercise 4. You will find examples in virtually every chapter, but you might want to go first to the sections on Newton’s 2nd Law, motion in viscous media, the charging of capacitors and inductors and radioactive decay. Exercise 5. Eqn. 10 separates immediately as dy y = ¡ q ( x ) dx: Exercise 6. a. Nonlinear, because of the y 2 factor in the second term. The equation is separable: dy y 2 = ¡ 2 xdx; with solution y ( x ) = 1 x 2 + C : b. Linear, inhomogeneous and separable. dy
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This note was uploaded on 07/13/2008 for the course PHY 201 taught by Professor Covatto during the Spring '08 term at ASU.

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FODE_1-10 - Tutorial FODE: Ordinary First-Order...

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